% ----------------------------------------------------------------
% AMS-LaTeX Paper ************************************************
% **** -----------------------------------------------------------
\documentclass[12pt]{amsart}


\usepackage{graphicx}
\usepackage{amssymb}
\usepackage{mathrsfs}
\usepackage{amsmath}
\usepackage[english]{babel}
\usepackage[all]{xy}
\usepackage[T1]{fontenc}
\usepackage[latin1]{inputenc}
\usepackage{newlfont}
%\usepackage{xypic}
\usepackage{tikz}

\newcommand{\np}{\not\in}
\newcommand{\intos}{\twoheadrightarrow}
\newcommand{\intoi}{\rightarrowtail}
\newcommand{\incl}{\hookrightarrow}
\newcommand{\p}{\mathfrak{p}}
\newcommand{\m}{\mathfrak{m}}
\newcommand{\n}{\mathfrak{n}}
\newcommand{\q}{\mathfrak{q}}
\newcommand{\am}{\mathfrak{a}}
\newcommand{\bm}{\mathfrak{b}}
\newcommand{\h}{\mathfrak{h}}
\newcommand{\ob}{\mathbf{Ob}}
\newcommand{\Fi}{\varphi}
\newcommand{\lam}{\lambda}
\newcommand{\PP}{\mathbf{P}}
\newcommand{\LL}{\mathscr{L}}
\newcommand{\Al}{\mathbf{A}}
\newcommand{\F}{\mathscr{F}}
\newcommand{\E}{\mathscr{E}}
\newcommand{\Pc}{\mathscr{P}}
\newcommand{\As}{\mathscr{A}}
\newcommand{\ZZ}{\mathscr{Z}}
\newcommand{\DD}{\mathscr{D}}
\newcommand{\G}{\mathscr{G}}
\newcommand{\HH}{\mathscr{H}}
\newcommand{\OO}{\mathscr{O}}
\newcommand{\JJ}{\mathscr{J}}
\newcommand{\OX}{\mathscr{O}_{X}}
\newcommand{\VV}{\mathscr{V}}
\newcommand{\Ms}{\widetilde{M}}
\newcommand{\Ns}{\widetilde{N}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\GG}{\mathbb{G}}
\newcommand{\Le}{\mathbb{L}}
\newcommand{\Ta}{\mathbb{T}}
\newcommand{\Cpl}{\mathbb{C}}
\newcommand{\RR}{\mathbb{R}}
\newcommand{\N}{\mathbb{N}}
\newcommand{\Fc}{\mathbf{F}}
\newcommand{\Aesp}{\mathbb{A}}
\newcommand{\restr}[1]{\big|_{#1}}
\newcommand{\cross}{\times}
\newcommand{\XF}{\Gamma(X,\F)}
\newcommand{\pt}{\mbox{\rm pt}}
\newcommand{\dual}{\vee}
\newcommand{\sU}{\mbox{\german U}}
\newcommand{\gm}{\mbox{\german m}}
\newcommand{\isom}{\simeq}
\newcommand{\req}{\thicksim}
\newcommand{\tensor}{\otimes}
\newcommand{\into}{\rightarrow}
\newcommand{\C}{\subset}
\newcommand{\V}{\cap}
\newcommand{\U}{\cup}
\newcommand{\iso}{\xrightarrow{\sim}}
\newcommand{\Gs}{\mathfrak{S}}
\newcommand{\llb}{\left[\kern-0.15em\left[}
\newcommand{\rrb}{\right]\kern-0.15em\right]}
\newcommand{\psus}[3]{#1_{#2},\dots,#1_{#3}}
\DeclareMathOperator{\Ob}{Ob}
\DeclareMathOperator{\Char}{char}
\DeclareMathOperator{\aut}{Aut}
\DeclareMathOperator{\Hc}{H}
\DeclareMathOperator{\End}{End}
\DeclareMathOperator{\GL}{GL}
\DeclareMathOperator{\slm}{SL}
\DeclareMathOperator{\supp}{supp}
\DeclareMathOperator{\Spec}{Spec}
\DeclareMathOperator{\Ext}{Ext}
\DeclareMathOperator{\Tor}{Tor}
\DeclareMathOperator{\Hom}{Hom}
\DeclareMathOperator{\Aut}{Aut}
\DeclareMathOperator{\proj}{Proj}
\DeclareMathOperator{\Pic}{Pic}
\DeclareMathOperator{\Img}{Im}
\DeclareMathOperator{\Coker}{Coker}
\DeclareMathOperator{\coim}{Coim}
\DeclareMathOperator{\id}{id}
\DeclareMathOperator{\Ker}{Ker}
\DeclareMathOperator{\Mod}{\mathbf{Mod}}
\DeclareMathOperator{\Coh}{\mathbf{Coh}}
\DeclareMathOperator{\Vgr}{\mathbf{VecGr}}
\DeclareMathOperator{\Vect}{\mathbf{Vect}}
\DeclareMathOperator{\SVect}{S-\mathbf{Vect}}
\DeclareMathOperator{\Corr}{\mathbf{Corr}}
\DeclareMathOperator{\cd}{cd}
\DeclareMathOperator{\ord}{ord}
\DeclareMathOperator{\depth}{depth}
\DeclareMathOperator{\trdeg}{trdeg}
\DeclareMathOperator{\Gal}{Gal}
\DeclareMathOperator{\rank}{rank}
\DeclareMathOperator{\car}{car}
\DeclareMathOperator{\rg}{rg}
\DeclareMathOperator{\Tr}{Tr}
\DeclareMathOperator{\sgn}{sgn}
\DeclareMathOperator{\Sym}{Sym}
\DeclareMathOperator{\ev}{ev}
\DeclareMathOperator{\coev}{coev}
\DeclareMathOperator{\length}{length}
\DeclareMathOperator{\Hilb}{Hilb}
\DeclareMathOperator{\Alb}{Alb}
\DeclareMathOperator{\rat}{rat}
\DeclareMathOperator{\num}{num}
\DeclareMathOperator{\alg}{alg}
\DeclareMathOperator{\nil}{nil}
\DeclareMathOperator{\cl}{cl}
\DeclareMathOperator{\pr}{pr}
\DeclareMathOperator{\op}{op}
\DeclareMathOperator{\et}{et}
\DeclareMathOperator{\eff}{eff}
\DeclareMathOperator{\kim}{kim}
\DeclareMathOperator{\Fr}{Fr}
\DeclareMathOperator{\AJ}{AJ}
\DeclareMathOperator{\CHM}{\text{\em CHM}}
\DeclareMathOperator{\NM}{\text{\em NM}}
\DeclareMathOperator{\CH}{\text{\em CH}}
\DeclareMathOperator{\un}{\mathbf{1}}
\DeclareMathOperator{\Ind}{\text{Ind}}
\DeclareMathOperator{\Res}{\text{Res}}
\DeclareMathOperator{\alb}{alb}
\DeclareMathOperator{\DR}{DR}
\DeclareMathOperator{\B}{B}
\DeclareMathOperator{\cris}{cris}
\DeclareMathOperator{\K3}{K3}
\DeclareMathOperator{\Div}{Div}
\DeclareMathOperator{\CaDiv}{CaDiv}
\DeclareMathOperator{\cyc}{cyc}
\DeclareMathOperator{\htt}{ht}
\DeclareMathOperator{\Cl}{Cl}
\DeclareMathOperator{\CaCl}{CaCl}
\DeclareMathOperator{\Span}{Span}
\DeclareMathOperator{\colim}{Colim}

% ----------------------------------------------------------------
\vfuzz2pt % Don't report over-full v-boxes if over-edge is small
\hfuzz2pt % Don't report over-full h-boxes if over-edge is small
% THEOREMS -------------------------------------------------------
\newtheorem{thm}{Theorem}[section]
\newtheorem{cor}[thm]{Corollary}
\newtheorem{lem}[thm]{Lemma}
\newtheorem{prop}[thm]{Proposition}
\theoremstyle{definition}
\newtheorem{defn}[thm]{Definition}
\theoremstyle{remark}
\newtheorem{rem}[thm]{Remark}
%\theoremstyle{example}
\newtheorem{exam}[thm]{Example}
\numberwithin{equation}{section}
% MATH -----------------------------------------------------------
\newcommand \lra {\longrightarrow}
\newcommand{\norm}[1]{\left\Vert#1\right\Vert}
\newcommand{\abs}[1]{\left\vert#1\right\vert}
\newcommand{\set}[1]{\left\{#1\right\}}
\newcommand{\Real}{\mathbb R}
\newcommand{\eps}{\varepsilon}
\newcommand{\To}{\longrightarrow}
\newcommand{\BX}{\mathbf{B}(X)}
\newcommand{\A}{\mathcal{A}}
% ----------------------------------------------------------------
\begin{document}

{\large \today}

\title{SEMINAR}%

%\author{Joe Palacios}%
%\address{}%
%\email{}%

%\thanks{}%
%\subjclass{}%
%\keywords{}%

%\date{}%
%\dedicatory{}%
%\commby{}%
% ----------------------------------------------------------------


\maketitle
% ----------------------------------------------------------------


\section*{Sheafification}

In this notes $\cal{C}$ is a category, $\tau$ is a Grothendieck pre-topology on $\cal{C}$, $\cal{A}$ is an abelian category.
 Let $$F:\cal{C}^{\op}\into \cal{A}\,.$$ be a presheaf, let $U$ be a object of $\cal{C}$ and  $\left\{f_i:U_i\into U\right\}_{i\in I}$ a covering of $U$. We have a sequence,
\[F(U)\stackrel{\rho}{\rightarrow}\prod_{i\in I}F(U_i)\;^{\stackrel{\rho_1}{\rightarrow}} _{\stackrel{\rightarrow}{\rho_2}}\;\prod_{i,j\in I}F(U_i\times_{U}U_j)\,,\]

 Let us recall the construction of $\rho,\rho_1$ and $\rho_2$. 

From the cartesian diagram,
 \[\xymatrix{U_i\times_{U} U_j\ar[r]\ar[d]& U_j\ar[d]^{f_j}\\U_i\ar[r]_{f_i}&U}\]
we get a commutative diagram,
\[\xymatrix{F(U_i\times_{U} U_j)& F(U_j)\ar[l]_{\;\rho_{2,i,j}}\\F(U_i)\ar[u]^{\rho_{1,i,j}}&F(U)\ar[u]_{f^*_j}\ar[l]^{f^*_i}}\,.\]
 Consider the product $\prod_{k\in I}F(U_k)$  and the projections $$\pi_i: \prod_{k\in I}F(U_k)\into F(U_i)$$ for every $i\in I$.
Since we have morphisms $f^*_i:F(U)\into F(U_i)$,  $i\in I$, by the universal property of the product, there exists a morphism  
$$\rho:F(U)\into\prod_{i\in I}F(U_i)$$ 
satisfying,
\begin{equation}\label{eq2078}
\pi_i\circ\rho=f^*_i
\end{equation}
for every $i\in I$. 

 Now, for every pair $(i,j)\in I^2$,  we can take the composition of $\pi_i:\prod_{k\in I}F(U_k)\into F(U_i)$ and $\rho_{1,i,j}:F(U_i)\into F(U_i\times_{U} U_j)$ and get a morphism,
$$\rho_{1,i,j}\circ \pi_i:\prod_{k\in I}F(U_k)\into F(U_i\times_{U} U_j)\,,$$
therefore, by the universal property of the product, there exists a unique morphism,
$$\rho_{1}:\prod_{k\in I}F(U_k)\into \prod_{i,j\in I}F(U_i\times_{U}U_j)$$
such that,
\begin{equation}\label{eq23}
\pi_{i,j}\circ\rho_{1}=\rho_{1,i,j}\circ \pi_i
\end{equation}
where,
 $$\pi_{i,j}:\prod_{i',j'\in I}F(U_{i'}\times_{U} U_{j'})\into F(U_i\times_{U} U_j)$$
are the canonical projections for indices $(i,j)\in I^2$.


Analogously,  for every pair $(i,j)\in I^2$, we take the composition of $\pi_j:\prod_{k\in I}F(U_k)\into F(U_j)$ and $\rho_{2,i,j}:F(U_j)\into F(U_i\times_{U} U_j)$ gives a morphism,
$$\rho_{2,i,j}\circ \pi_j:\prod_{k\in I}F(U_k)\into F(U_i\times_{U} U_j)$$
hence, by the universal property of the product, there exists a unique morphism,
$$\rho_{2}:\prod_{k\in I}F(U_k)\into \prod_{i,j\in I}F(U_i\times_{U} U_j)$$
such that,
\begin{equation}\label{eq24}
\pi_{i,j}\circ\rho_{2}=\rho_{2,i,j}\circ \pi_j
\end{equation}

\begin{lem}\label{one12}
We have $\rho_1\circ\rho=\rho_2\circ\rho$.
\end{lem}
{\em Proof:} From the commutative diagram,
\[\xymatrix{F(U_i\times_{U} U_j)& F(U_j)\ar[l]_{\;\rho_{2,i,j}}\\F(U_i)\ar[u]^{\rho_{1,i,j}}&F(U)\ar[u]_{f^*_j}\ar[l]^{f^*_i}}\,.\]
we get $\rho_{1,i,j}\circ f^*_i=\rho_{2,i,j}\circ f^*_j$. 
On the other hand, from (\ref{eq2078}), we have $\pi_i\circ \rho=f^*_i$ for every $i\in I$. Then, from (\ref{eq23}) and (\ref{eq24}), we have, 
\begin{equation}\label{}
\begin{split}
(\pi_{i,j}\circ\rho_{1})\circ\rho&=(\rho_{1,i,j}\circ\pi_i)\circ\rho=\rho_{1,i,j}\circ(\pi_i\circ\rho)=\rho_{1,i,j}\circ f^*_i=\rho_{2,i,j}\circ f^*_j\\&
=\rho_{2,i,j}\circ(\pi_j\circ \rho)=(\rho_{2,i,j}\circ\pi_j)\circ \rho=(\pi_{i,j}\circ\rho_{2})\circ\rho
\end{split}
\end{equation}
thus we have $\pi_{i,j}\circ(\rho_{1}\circ\rho)=\pi_{i,j}\circ(\rho_{2}\circ\rho)$. Therefore, using the universal property of $\prod_{i,j\in I}F(U_i\times_{U}U_j)$, we get
$$\rho_{1}\circ\rho=\rho_{2}\circ\rho\,.$$ 
 
\hfill $\Box$

\begin{lem} Let $F$ be a presheaf $\cal{C}^{\op}\into \cal{A}$. 
\begin{enumerate}
\item[(a)] $F^{+}$ is separated.
\item[(b)] If $F$ is separated, then $F\into F^{+}$ is a monomorphism.
\end{enumerate}
\end{lem}
The proof of this lemma is given in other manuscript. 

Let us concentrate in the following:

\begin{prop}
If $F$ is a separated presheaf, then $F^{+}$ is a sheaf.
\end{prop}
{\em Proof:}  Let $\{f_i:U_i\into U\}_{i\in I}$ be a covering of $U$. We must prove that following sequence
\[0\into F^{+}(U)\stackrel{\rho^+}{\rightarrow} \prod_{i\in I} F^{+}(U_i)\stackrel{\rho^+_1-\rho^+_2}{\rightarrow}\prod_{i,i'\in I}F^{+}(U_i\times_U U_{i'})\]
 is exact.

 By the previous lemma we have that $F^{+}$ is separated, in particular the map $F^{+}(U)\into \prod_{i\in I} F^{+}(U_i)$ is a monomorphism.
 We shall prove that $\Img\rho^+=\Ker(\rho^+_1-\rho^+_2)$.

 By lemma \ref{one12}, we have $(\rho^+_1-\rho^+_2)\circ \rho^+=0$,  then $\Img\rho^+\C\Ker(\rho^+_1-\rho^+_2)$. We have only to show that $\Ker(\rho^+_1-\rho^+_2)\C\Img\rho^+$. Let $(x_i)_{i\in I}$ be an arbitrary element of $\Ker(\rho^+_1-\rho^+_2)$. We shall find an element of $x\in F^+(U)$ such that $$\rho^+(x)=(x_i)_{i\in I}\,.$$
 Since $\Ker(\rho^+_1-\rho^+_2)\C \prod_{i\in I} F^{+}(U_i)$, then $(x_i)_{i\in I}$ is an element of $\prod_{i\in I} F^{+}(U_i)$, where $x_i\in F^+(U_i)$ for every $i\in I$. By definition, we have $$F^+(U_i)=\colim_{\cal{J}_{U_i}}H^0_{U_i}(-,F)\,.$$
Then for each $i\in I$, there exist a covering $\cal{V}_i=\left\{V_{ik}\into U_i\right\}$ in $\cal{J}_{U_i}$, a morphism $\mu_{\cal{V}_i}:H^0_{U_i}(\cal{V}_i,F)\into F^+(U_i)$ and an element $y_i\in H^0_{U_i}(\cal{V}_i,F)$ such that, $$\mu_{\cal{V}_i}(y_i)=x_i\,.$$
By definition of $H^0_{U_i}(\cal{V}_i,F)$, it is a subset of $\prod_{k} F(V_{ik})$ then $y_i$ is a collection $(y_{ik})_{k}\in \prod_{k} F(V_{ik})$. Thus the collection $(y_{ik})_{ik}\in\prod_{ik} F(V_{ik})$ determines the element $(x_i)_{i}\in\prod_{i\in I} F^{+}(U_i)$.

Let $\cal{V}$ be the covering $\left\{V_{ik}\into U_i\into U\right\}\in \cal{J}_U$. In particular, we have an exact sequence,

\[0\into H^0_{U}(\cal{V},F)\into \prod_{ik} F(V_{ik})\stackrel{\rho_{1}-\rho_{2}}{\longrightarrow} \prod_{iki'k'} F(V_{ik}\times_{U}V_{i'k'})\]

\vspace{.3cm}
{\bf CLAIM 1: } The element $(y_{ik})_{ik}\in\prod_{ik} F(V_{ik})$ lies in $H^0_{U}(\cal{V},F)$.

\vspace{.3cm}

To prove the above claim, it suffices to show that $$(y_{ik})_{ik}\in\Ker(\rho_{1}-\rho_{2})\,.$$

Since $F\into F^+$ is monomorphism by the previous lemma, in particular we have a monomorphism $F(V_{ik}\times_{U}V_{i'k'})\into F^+(V_{ik}\times_{U}V_{i'k'})$, hence taking the product of all this morphisms, we get a monomorphism
$$\prod_{iki'k'} F(V_{ik}\times_{U}V_{i'k'})\into \prod_{iki'k'} F^+(V_{ik}\times_{U}V_{i'k'})\,.$$
For each $i$, consider the exact sequence,
\[0\into H^0_{U_i}(\cal{V}_i,F)\into \prod_{k} F(V_{ik})\into \prod_{kk'} F(V_{ik}\times_{U_i}V_{ik'})\]
then we get a natural monomorphism $$\prod_{i}H^0_{U_i}(\cal{V}_i,F)\into \prod_{ik} F(V_{ik})\,.$$
On the other hand, the natural morphisms $$F^{+}(U_i\times_U U_{i'})\into F^+(V_{ik}\times_{U}V_{i'k'})$$
 induce a morphism,
$$\prod_{i,i'\in I}F^{+}(U_i\times_U U_{i'})\into\prod_{iki'k'} F^+(V_{ik}\times_{U}V_{i'k'})\,.$$ 
Considering the above constructions, we have the following diagram,
\[\xymatrix{F^{+}(U)\ar[r]&\prod_{i} F^{+}(U_i)\ar[r]& \prod_{i,i'\in I}F^{+}(U_i\times_U U_{i'})\ar[r]& \prod_{iki'k'} F^+(V_{ik}\times_{U}V_{i'k'})\\
&\prod_{i}H^0_{U_i}(\cal{V}_i,F)\ar[u]\ar[d]& &\\
H^0_{U}(\cal{V},F)\ar[r]&\prod_{ik} F(V_{ik})\ar[rr]^{\rho_1-\rho_2}&&\prod_{iki'k'} F(V_{ik}\times_{U}V_{i'k'})\ar[uu]}\] 
Since the image of the element $(y_{ik})_{ik}\in \prod_{i}H^0_{U_i}(\cal{V}_i,F)$ by the composition
\[\prod_{i}H^0_{U_i}(\cal{V}_i,F)\into\prod_{i} F^{+}(U_i)\into \prod_{i,i'\in I}F^{+}(U_i\times_U U_{i'})\into\prod_{iki'k'} F^+(V_{ik}\times_{U}V_{i'k'}) \]
is zero and the vertical arrow on the right-hand-side is a monomorphism, to prove that $(y_{ik})_{ik}\in\Ker(\rho_{1}-\rho_{2})$, it suffices to show that the above diagram is commutative. 

For a fixed index $i\in I$, notice that the following  diagram 
\[\xymatrix{U_i& V_{ik}\ar[l]\\
U_i\times_U U_i\ar[u]& V_{ik}\times _U V_{ik'}\ar[u]\ar[l]}\]
is commutative for any couple of indices $k,k'$, moreover, they induce the following diagram 

\[\xymatrix{F(U_i)\ar[r]\ar[d]& F(V_{ik})\ar[d]\\
F(U_i\times_U U_i)\ar[r]&F(V_{ik}\times _U V_{ik'})}\]
hence the diagram, 
\[\xymatrix{F(U_i)\ar[r]\ar[d]&\prod_{k}F(V_{ik})\ar[d]\\
F(U_i\times_U U_i)\ar[r]&\prod_{kk'}F(V_{ik}\times _U V_{ik'})}\]
Since $\cal{V}_i=\{V_{ik}\into U_i\}_k$ is a covering of $U_i$, there exists a morphism $H^0_{U_i}(\{U_i\into U_i\}, F)\into H^0_{U_i}(\cal{V}_i, F)$
making the diagram 
\[\xymatrix{0\ar[d]&0\ar[d]\\
H^0_{U_i}(\{U_i\into U_i\}, F)\ar[r]\ar[d]&H^0_{U_i}(\cal{V}_i, F)\ar[d]\\
F(U_i)\ar[r]\ar[d]&\prod_{k}F(V_{ik})\ar[d]\\
F(U_i\times_U U_i)\ar[r]&\prod_{kk'}F(V_{ik}\times _U V_{ik'})}\]
commutative, where $H^0_{U_i}(\{U_i\into U_i\}, F)$ is nothing else that $F(U_i)$; thus we have a morphism $F(U_i)\into H^0_{U_i}(\cal{V}_i, F)$, and hence we get a morphism $$\prod_{i}F(U_i)\into \prod_{i} H^0_{U_i}(\cal{V}_i, F)\,.$$
Considering this morphism and the natural morphisms, we have the following diagram,
\[\xymatrix{\prod_{i} F^{+}(U_i)\ar[rr]^{\rho^+_1-\rho^+_2}&& \prod_{i,i'}F^{+}(U_i\times_U U_{i'})\ar[rr]&& \prod_{iki'k'} F^+(V_{ik}\times_{U}V_{i'k'})\\
&\ar @{} [dl] |{\text{(B)}}&&\text{(A)}&\\
\prod_{i}H^0_{U_i}(\cal{V}_i,F)\ar[uu]\ar[dd]&&\prod_{i}F(U_i)\ar[uull]\ar[lldd]\ar[ll]\ar[rrdd]&&\\
&\ar @{} [ul] |{\text{(C)}}&\text{(D)}&\\
\prod_{ik} F(V_{ik})\ar[rrrr]^{\rho_1-\rho_2}&&&&\prod_{iki'k'} F(V_{ik}\times_{U}V_{i'k'})\ar[uuuu]}\]
We shall show that every diagram $(A)$, $(B)$, $(C)$ and $(D)$ is commutative. 

\vspace{.3cm}

{\bf Diagram (A) is commutative:}

\vspace{.3cm}  
First of all notice that the commutative diagram, 
\[\xymatrix{U_i& U_i\times_{U}U_{i'}\ar[l]\\
& V_{ik}\times _U V_{i'k'}\ar[u]\ar[lu]}\]
induce a commutative diagram,
  \[\xymatrix{F^+(U_i)\ar[r]\ar[rd]& F^+(U_i\times_{U}U_{i'})\ar[d]\\
& F^+(V_{ik}\times _U V_{i'k'})}\]
hence we have the commutative diagram, 
  \[\xymatrix{\prod_{i}F^+(U_i)\ar[r]\ar[rd]&\prod_{ii'}F^+(U_i\times_{U}U_{i'})\ar[d]\\
& \prod_{iki'k'}F^+(V_{ik}\times _U V_{i'k'})}\]
Then, to prove that $(A)$ is a commutative diagram, it suffices that the following diagram 
 \[\xymatrix{\prod_{i}F^+(U_i)\ar[r]\ar[d]&\prod_{iki'k'}F^+(V_{ik}\times _U V_{i'k'})\ar[d]\\
\prod_{i}F(U_i)\ar[r]&\prod_{iki'k'}F(V_{ik}\times _U V_{i'k'})}\]
is commutative. The commutativity of this diagram follows immediately since the morphism $F\into F^+$ is a natural transformation and we have a commutative diagram,
 \[\xymatrix{F^+(U_i)\ar[r]\ar[d]&F^+(V_{ik}\times _U V_{i'k'})\ar[d]\\
F(U_i)\ar[r]&F(V_{ik}\times _U V_{i'k'})}\]

\vspace{.3cm}

{\bf Diagrams (B) and (C) are commutative:}

\vspace{.3cm}  
 
We have already constructed the following commutative diagram,   

\[\xymatrix{0\ar[d]&0\ar[d]\\
H^0_{U_i}(\{U_i\into U_i\}, F)\ar[r]\ar@{=}[d]&H^0_{U_i}(\cal{V}_i, F)\ar[d]\\
F(U_i)\ar[r]&\prod_{k}F(V_{ik})}\]
This gives us a commutative triangle, 

 \[\xymatrix{H^0_{U_i}(\cal{V}_i, F)\ar[d]&F(U_i)\ar[ld]\ar[l]\\
\prod_{k}F(V_{ik})&}\,,\]
hence a commutative diagram, 
 \[\xymatrix{\prod_{i}H^0_{U_i}(\cal{V}_i, F)\ar[d]&\prod_{i}F(U_i)\ar[ld]\ar[l]\\
\prod_{ik}F(V_{ik})&}\,,\]
this is the diagram $(C)$. On another hand, by definition of colimit, we have a commutative triangle, 
 \[\xymatrix{F^+(U_i)&\\
H^0_{U_i}(\cal{V}_i,F)\ar[u]&H^0_{U_i}(\left\{U_i\into U_i\right\},F)\ar[ul]\ar[l]}\]
where $H^0_{U_i}(\left\{U_i\into U_i\right\},F)$ is $F(U_i)$, hence taking product, we get a commutative diagram, 
\[\xymatrix{\prod_{i}F^+(U_i)&\\
\prod_{i}H^0_{U_i}(\cal{V}_i,F)\ar[u]&\prod_{i} F(U_i)\ar[ul]\ar[l]}\]
this is the diagram $(C)$. 

\vspace{.3cm}

{\bf Diagram (D) is commutative:}

\vspace{.3cm} 
Notice that the diagram 
\[\xymatrix{&\prod_{i}F(U_i)\ar[rd]\ar[ld]&\\
\prod_{ik} F(V_{ik})\ar[rr]^{\rho_1-\rho_2}&&\prod_{iki'k'} F(V_{ik}\times_{U}V_{i'k'})}\]
is commutative, because the morphism $$\prod_{i}F(U_i)\into\prod_{iki'k'} F(V_{ik}\times_{U}V_{i'k'})$$
is taken as the composition 
\[\prod_{i}F(U_i)\into \prod_{ik} F(V_{ik})\stackrel{\rho_1-\rho_2}{\rightarrow} \prod_{iki'k'} F(V_{ik}\times_{U}V_{i'k'})\]
This proves the CLAIM 1, which states that $(y_{ik})_{ik}\in H^0_{U}(\cal{V},F)$ .

To finish the proof of the proposition we need only to prove the following:


\vspace{.3cm}
{\bf CLAIM 2: } The following diagram,  
\[\xymatrix{F^{+}(U)\ar[r]&\prod_{i} F^{+}(U_i)\\
H^0_{U}(\cal{V},F)\ar[r]\ar[u]&\prod_{i}H^0_{U_i}(\cal{V}_i,F)\ar[u]}\] 
is commutative. 

\vspace{.3cm}
To prove that above diagram is commutative, it suffices that following diagram is commutative, 
\[\xymatrix{F^{+}(U)\ar[r]&F^{+}(U_i)\\
H^0_{U}(\cal{V},F)\ar[r]\ar[u]&H^0_{U_i}(\cal{V}_i,F)\ar[u]}\]
Let $\cal{W}_i$ be the covering $\left\{V_{ik}\times_U U_i\into U_i\right\}_k$ of $U_i$, formed by projections. 
We recall that, by construction, the morphism $F^{+}(U)\into F^{+}(U_i)$ has the property that the following diagram, 
\[\xymatrix{F^{+}(U)\ar[r]&F^{+}(U_i)\\
H^0_{U}(\cal{V},F)\ar[r]\ar[u]\ar[ur]&H^0_{U_i}(\cal{W}_i,F)\ar[u]}\]   
is commutative. 
Notice that, on the other hand, that we have a commutative diagram, 
 \[\xymatrix{H^0_{U}(\cal{V},F)\ar[r]\ar[d]&H^0_{U_i}(\cal{V}_i,F)\ar[r]\ar[d]&H^0_{U_i}(\cal{W}_i,F)\ar[d]\\
\prod_{ik} F(V_{ik})\ar[r]\ar@<2pt>[d]\ar@<-2pt>[d]&\prod_{k} F(V_{ik})\ar[r]\ar@<2pt>[d]\ar@<-2pt>[d]&\prod_{k} F(V_{ik}\times_U U_i)\ar@<2pt>[d]\ar@<-2pt>[d]\\
&&
}\] 
Then the diagram
 \[\xymatrix{H^0_{U}(\cal{V},F)\ar[r]\ar@/_1pc/[rr]&H^0_{U_i}(\cal{V}_i,F)\ar[r]&H^0_{U_i}(\cal{W}_i,F)
}\]
is commutative. Thus, we have the following diagram,
\[\xymatrix{F^{+}(U)\ar[r]&F^{+}(U_i)&\\
H^0_{U}(\cal{V},F)\ar[r]\ar[u]\ar[ur]\ar@/_1pc/[rr]&H^0_{U_i}(\cal{V}_i,F)\ar[u]\ar[r]&H^0_{U_i}(\cal{W}_i,F)\ar[ul]}\]  
Finally, notice that to prove that the diagram
\[\xymatrix{F^{+}(U)\ar[r]&F^{+}(U_i)\\
H^0_{U}(\cal{V},F)\ar[r]\ar[u]\ar[ur]&H^0_{U_i}(\cal{V}_i,F)\ar[u]}\]
is commutative, we have only to check that the triangle,
\[\xymatrix{F^{+}(U_i)&\\
H^0_{U_i}(\cal{V}_i,F)\ar[r]\ar[u]&H^0_{U_i}(\cal{W}_i,F)\ar[ul]}\]
is commutative, but it is commutative by definition of colimit defining $F^{+}(U_i)$. 
This proves the CLAIM 2 and the proposition.

\hfill $\Box$


%\begin{equation}\label{rel123}
%\begin{split}
%v.
%\end{split}
 %\end{equation}


\begin{thebibliography}{99}

\bibitem[{MHT}]{MHT}
  Dundas,Levine,Ostvaer, Rondings, Voevodsky.
  \newblock \textit{Motivic Homotopy Theory}.
  \newblock Springer (2002).
\bibitem[{Tam}]{Tam}
  G. Tamme.
  \newblock \textit{Introduction to Etale Cohomology}.
  \newblock Springer-Verlag (1991).
\end{thebibliography}
\end{document}
% ----------------------------------------------------------------
